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If $\pi_1$ and $\pi_2$ are two smooth admissible representations of $\operatorname{GL}_2(\mathbb{Q}_p)$ over $\overline{\mathbb{F}}_p$ with central characters. I want to prove that if $\pi_1$ has central character $\phi$ and $\operatorname{Ext}^i(\pi_1, \pi_2)\neq0$ for some $i>0$, then $\phi$ is also the central character of $\pi_2$.

Based on my previous question, we know that this holds for $i=1$. For $i>1$, if $\operatorname{Ext}^i(\pi_1, \pi_2)\neq0$, then we have a non-trivial extension $0\to\pi_2\to V_1\to\dotsb\to V_i\to\pi_1\to0$. In this case, non-triviality means that there is no morphism from $0\to\pi_2\to V_1\to\dotsb\to V_i\to\pi_1\to0$ to $0\to\pi_2\to \pi_2\to0\to\dotsb\to0\to \pi_1\to\pi_1\to0$.

I think this result is well-known to experts, could someone explain how to prove this or give a reference?

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    $\begingroup$ Without rechecking the details I think you can argue that the centre must act on all the Ext groups by both the central character of $\pi_1$ and the central character of $\pi_2$. If the Ext group is Nonzero then the central characters must be equal. The argument is by 'functoriality' $\endgroup$
    – Simon Wadsley
    Commented Oct 25, 2023 at 17:17
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    $\begingroup$ The exactly same argument I wrote also applies here. The smooth representations $V_i$ split as a representation of the center $\mathbb Q_p^\times\hookrightarrow\mathrm{GL}_2(\mathbb Q_p)$. You can use the splitting to prove the sequence is trivial. $\endgroup$
    – Kenta Suzuki
    Commented Oct 25, 2023 at 18:47
  • $\begingroup$ @KentaSuzuki's argument referenced above. $\endgroup$
    – LSpice
    Commented Oct 25, 2023 at 21:38

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